Question

Two mine haulage roadways A and B are having similar cross-sectional areas, surface characteristics, but differ in lengths. The length road way A is 100 m and that of road way B is 200 m. The ratio of their resistances R\(_A\): R\(_B\) is

• A. 1 : 2
• B. 1 : 4
• C. 1 : 3
• D. 1 : 9

Hint:

For airways with the same shape, size, and lining, the resistance is simply proportional to the length.
If you double the length, you double the resistance.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the ratio of aerodynamic resistance of two mine roadways that are identical except for their lengths.

Step 2: Key Formula or Approach:
The aerodynamic resistance (R) of a mine airway is given by Atkinson's formula: \[ R = \frac{k L P}{A^3} \] where: - \(k\) = friction factor (depends on surface characteristics) - \(L\) = length of the roadway - \(P\) = perimeter of the roadway's cross-section - \(A\) = cross-sectional area of the roadway

Step 3: Detailed Explanation:
The problem states that the two roadways have: - Similar cross-sectional areas (\(A_A = A_B\)). - Similar surface characteristics (\(k_A = k_B\)). - Similar area implies a similar perimeter (\(P_A = P_B\)).
This means that all the terms in the resistance formula are constant except for the length (L).
Therefore, the resistance (R) is directly proportional to the length (L).
\[ R \propto L \] We can find the ratio of the resistances by taking the ratio of the lengths: \[ \frac{R_A}{R_B} = \frac{L_A}{L_B} \] Given \( L_A = 100 \, \text{m} \) and \( L_B = 200 \, \text{m} \).
\[ \frac{R_A}{R_B} = \frac{100}{200} = \frac{1}{2} \] The ratio R\(_A\): R\(_B\) is 1 : 2.

Step 4: Final Answer:
The ratio of their resistances R\(_A\): R\(_B\) is 1 : 2.
This corresponds to option (A).